Answer
$= \frac{2(3g+1)}{g(3g-1)}, with$ $g\ne\frac{1}{3}$
Work Step by Step
Given : $\frac{\frac{g+2}{3g-1}}{\frac{g^{2}+2g}{6g+2}}$
This becomes : $\frac{g+2}{3g-1} \times \frac{2\times(3g+1)}{g(g+2)}$
$= \frac{2(3g+1)}{g(3g-1)}$
(After dividing out the common factor (g+2))
